Finding the general solution to a differential equation

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Homework Help Overview

The discussion revolves around finding the general solution to a second-order linear differential equation with constant coefficients, specifically involving a non-homogeneous term that includes exponential and trigonometric functions.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve the homogeneous equation and proposes a particular solution, but encounters difficulties. Some participants suggest modifying the form of the particular solution due to the presence of repeated terms in the non-homogeneous part.

Discussion Status

The discussion is ongoing, with participants exploring different forms of the particular solution. There is a mix of attempts and suggestions, but no consensus has been reached regarding the correct approach.

Contextual Notes

Participants are considering the implications of the non-homogeneous term and the necessity of adjusting the assumed form of the particular solution based on the characteristics of the complementary solution.

clarineterr
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Homework Statement


[tex]\frac{d^{2}y}{dt}[/tex] +4[tex]\frac{dy}{dt}[/tex]+20y=e[tex]^{-2t}[/tex](sin4t+cos4t)

Homework Equations





The Attempt at a Solution



The solution to the homogeneous equation: [tex]\frac{d^{2}y}{dt}[/tex] +4[tex]\frac{dy}{dt}[/tex]+20y=0 is

y= k1e[tex]^{-2t}[/tex]cos4t +k2e[tex]^{-2t}[/tex]sin4t

Then I guessed ae[tex]^{-2+4i}[/tex] as a possible solution and it didn't work, and that's where I'm stuck.
 
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Since the complementary solution yc= e-2t(Acos(4t) + Bsin(4t)) is repeated in the non-homogeneous term, for your particular solution try

yp = te-2t(Ccos(4t) + Dsin(4t))
 
Nope...didn't work. :(
 
clarineterr said:
Nope...didn't work. :(

Yes, it does work. Check your work or show it here if you can't find your error.
 

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